# Moments problemWatch

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#1
Hello everybody,

I'm having trouble with labelling the forces (part a of the document) - I know that the skip is exerting a force n the beam, but I don't know of any other forces - I know about the centre of mass as well.

So which way would the force from the skip be acting, and are there only these two forces acting on the beam?
0
2 years ago
#2
(Original post by Electrogeek)
Hello everybody,

I'm having trouble with labelling the forces (part a of the document) - I know that the skip is exerting a force n the beam, but I don't know of any other forces - I know about the centre of mass as well.

So which way would the force from the skip be acting, and are there only these two forces acting on the beam?
At A there is a horizontal reaction force to the right, and at B there is a reaction force pointing upwards, perpendicular to the beam. Then it's just the CoM.
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#3
(Original post by RDKGames)
At A there is a horizontal reaction force to the right, and at B there is a reaction force pointing upwards, perpendicular to the beam. Then it's just the CoM.
Thanks for the help - would S and the sine component of force R add together to make the weight of the rod?

Finally, am I right in saying that the angle of R is the same as that in part b (Or equivalent to)?
0
2 years ago
#4
(Original post by Electrogeek)
Thanks for the help - would S and the sine component of force R add together to make the weight of the rod?

Finally, am I right in saying that the angle of R is the same as that in part b (Or equivalent to)?
Can't say for certain because I do not know which one's your R and which one's your S.

When considering the system overall, the horizontal force at A is perpendicular to the direction of the weight so you wouldn't include that force. You need to take moments about either A or B instead, and preferably parallel/perp to the beam.
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#5
(Original post by RDKGames)
Can't say for certain because I do not know which one's your R and which one's your S.

When considering the system overall, the horizontal force at A is perpendicular to the direction of the weight so you wouldn't include that force. You need to take moments about either A or B instead, and preferably parallel/perp to the beam.
My R is the force at A, and S is the force at B
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2 years ago
#6
So the moment of the weight (about point A) would be 500cos22 * 2.5?
Then what would you have to do?
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